How can I prove that $\mathbb{E}[g^{2k}] = (2k-1)!!$

Question

Let $k$ be a non-negative integer such that $k \geq 0$. Let $g$ ~ $N(0,1)$. From an earlier proof, I know that $\mathbb{E}gF(g) = \mathbb{E}F'(g)$ with $F: \mathbb{R} \rightarrow \mathbb{R}$. I also know that $\Gamma(z) = \int_{0}^{\infty} x^{z-1}e^{-x}dx$. With this, I have the equation: $$\mathbb{E}g^{2k} = \int_{-\infty}^{\infty} \frac{2kx^{2k-1}}{\sqrt{2\pi}}e^{-x^2/2}dx$$ $$\mathbb{E}x^{2k} = \frac{2k}{\sqrt{2\pi}}\int_{0}^{\infty} \frac{x^{2k-1}}{\sqrt{2\pi}} dx$$ and I am not sure where to go from here to show that $\mathbb{E}g^{2k} = (2k-1)!!$. Any insight would be appreciated.

Answer 1

Your second integral cannot be correct because it diverges. $$ \begin{align} \mathbb{E}\!\left[g^{2k}\right] &=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty x^{2k}e^{-x^2/2}\,\mathrm{d}x\tag1\\ &=\frac{2^{k+1/2}}{\sqrt{2\pi}}\int_0^\infty x^{2k-1}e^{-x^2}\,\mathrm{d}x^2\tag2\\ &=\frac{2^k}{\sqrt\pi}\int_0^\infty x^{k-1/2}e^{-x}\,\mathrm{d}x\tag3\\ &=2^k\frac{\Gamma(k+1/2)}{\Gamma(1/2)}\tag4\\[6pt] &=(2k-1)!!\tag5 \end{align} $$ Explanation:
$(1)$: expected value of $x^{2k}$ given $N(0,1)$
$(2)$: substitute $x\mapsto x\sqrt2$ and use symmetry
$(3)$: substitute $x\mapsto x^{1/2}$
$(4)$: apply the Gamma Function integral
$(5)$: use that $2^k\Gamma(k+1/2)=(2k-1)2^{k-1}\Gamma(k-1/2)$